Question

Please help, help!!!!!!! Using the data given in the table, answer the question

Tabulate Imedy 506 Mean 3.0345128744 Std Dev 0.4087568504 Min 1.6094379124 Max 3.9120230054 Range 2.302585093 Variance 0.1670821628 Std Err 0.0181714585 Median Interquartile Range 3.0540011817 0.3886210952 Quantiles0.25 1.6094379124 Quantiles0.5 1.6700687591 Quantiles0.75 1.8172874839

(a) Construct a 95% confidence interval for lmedv and test the hypothesis H0: mean lmedv = 3 vs. H1: mean lmedv > 3. What is the test statistic value, p-value and your conclusion of the test at α = 0.05? (Note that the confidence interval is two-sided while the test is one-sided.) lmdev being median value of owner-occupied homes in $1000's

(b) Let p = proportion of lmedv > 3.02 (i.e., lmedv2=1). Test H0: p = 0.5 vs. H1: p > 0.5 at α = 0.05.

Answer 1

a)

Mean(x) = 3.034

n = 506

Std Dev(s) = 0.4087

Alpha = 0.05

Z_Critical = 1.96

Hence,

95% CI = Mean +/- Z_Critical * Std Dev / n^1/2 = 3.034 +/- 1.96*0.4087/506^1/2

= {2.998,3.070}

Null and Alternate Hypothesis

H₀: µ = 3

H_a: µ > 3

Test Statistic

t = (x - µ₀)/(s/n^1/2) = (3.034-3) / ( 0.4087/506^1/2) = 1.87

p-value = TDIST(1.87,506-1,1) = 0.030939

Result

Since the p-value is less than 0.05, we reject the null hypothesis.

b) Data is wrong as proportion is given to be 3.02 (It can never be greater than 1)